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How to solve for time given distance and acceleration using calculus?

Suppose you drop a ball from the top of a building that is 100 feet tall. When you drop the ball, you do not give it any initial velocity. Using calculus, how long does it take (in seconds) for the ball to hit the ground? Assume the acceleration due to gravity is 32.2 ft/sec^2

I'm not sure how to start. I know that acceleration is the derivative of velocity, velocity is the derivative of position, and so forth.

7 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Since this is a Calculus problem, we will start with

    dV/dt = a

    where

    dV/dt = acceleration of the ball as it is dropped from top of the building

    a = acceleration due to gravity = 32.2 ft/sec^2 (constant)

    The above equation then becomes

    dV/dt = 32.2

    V = ⌠32.2(dt)

    and integrating the above,

    V = 32.2t + C

    where

    C = integration constant

    Since the ball was dropped at t = 0, then

    0 = 0 + C and C = 0, hence the equation for V becomes

    V = 32.2t

    Since V = dS/dt, then

    dS/dt = 32.2t

    and

    S = ⌠(32.2t)dt

    Integrating the above,

    S = (1/2)(32.2t^2) + C

    and when t = 0, the ball has not moved yet, so

    0 = 0 + C

    and the above equation for "S" simplifies to

    S = 16.1t^2

    When S = 100 (distance travelled by the ball),

    100 = 16.1t^2

    and solving for "t"

    t = 2.49 sec

    Hope this helps.

  • vij
    Lv 6
    1 decade ago

    D= Vi(t) + 1/2 a(t^2) In your question D = 100 ft, Vi = 0,

    a = 32.2 ft/ sec^2

    Sub the values in the equation and solve for t.

    100 = 0(t) + 1/2(32.2) t^2

    100 = 16.1 x t^2

    100/ 16.1 = t^2

    6.2 = t^2

    2.5 = t

    So ball takes 2.5 sec to hit the ground.

  • 1 decade ago

    Well, the position equation for this is a*t^2 + v*t - h0, where h0 is the initial height, a is acceleration, and v is initial velocity. So, this position equation is 32.2*t^2 - 100. You set this equal to zero because you want the time that the height (or position) is 0 ft off the ground. So...

    0 = 32.2*t^2 - 100

    100 = 32.2*t^2 Divide by 32.2, take the (positive) square root, and you have the time when it hits the ground

  • Anonymous
    5 years ago

    Hi he.goes.the.distance Use the formula v^-u^2 = -2as (=2as when it is acceleration) v= 0 a = 16 s = 200 o^2 - u^2 = 2*16*200 u = 80 ft/sec Shy

  • 1 decade ago

    a = dv/dt = -32.2 ft/s²

    so integrate

    v = -32.2ft/s² t + v(0)

    but you said v(0) = 0 so

    v = (-32.2ft/s²) t

    but v = ds/dt

    so

    ds/dt = (-32.2ft/s²) t

    s = (-32.2ft/s²)(1/2) t² + s(0)

    so you want to know when s = 0 (final value on ground) when s(0) (initial position)= 100ft

    0 = (-16.1ft/s²) t² + 100ft

    Solve for t!

    [-100/-16.1]s² = t²

    6.2 s² = t²

    t = 2.489 sec

  • 1 decade ago

    Hint: if you drop a ball regardless of height, what is your initial velocity?

    s = -16.1t^2 + 100

    - 100 = -16.1t^2

    100 / 16.1 = t^2

    10 / sqrt(16.1) = t

  • 1 decade ago

    Use the equation:

    distance = a*t^2 / 2 = g*t^2 / 2

    t = √(2*distance/g)

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